Question

Решите систему неравенств:
x-2x-24 ≤0
4-3(x+1) > 5x

Answer 1

$\displaystyle\bf\\\left \{ {{ {x}^{2} - 2x - 24 \leqslant 0 } \atop {4 - 3(x + 1) > 5x }} \right. \\ \\ 1) \: {x}^{2} - 2x - 2 \leqslant 0 \\ {x}^{2} - 2x - 24 = 0 \\ po \: \: \: teoreme \: \: \: vieta \\ {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c \\ \\ x_{1}  +  x_{2} = 2 \\ x_{1} x_{2} =  - 24 \\ x_{1} =  6\\ x_{2} = - 4 \\ \\ {ax}^{2} + bx + c = a(x - x_{1})(x - x_{2}) \\ {x}^{2} - 2x - 24 = (x - 6)(x + 4) \\ \\ (x - 6)(x + 4) \leqslant 0 \\ + + + [ - 4] - - - [6] + + + \\ - 4 \leqslant x \leqslant 6 \\ \\ 2) \: 4 - 3(x + 1) > 5x \\ 4 - 3x - 3 - 5x > 0 \\ - 8x > - 1 \\ x < 0.125 \\ \displaystyle\bf\\\left \{ {{ - 4 \leqslant x \leqslant 6} \atop {x < 0.125 }} \right. \\ \\ x \: \epsilon \: [ - 4; \: 0.125)$